Question

could you help me with this limit problem :D ?

Answer 1

can you use L'Hopitals rule?

do you know derivatives?

Answer 2

hmm no yet, we are only allowed to use algebra techniques :D

Answer 3

ok then, hold on let me see if i can work it out

the idea is to cancel out the "x-8" in denominator so its not dividing by zero

Answer 4

yeah

Answer 5

if you don't like the cube root of x replace it with u
then rationalize the numerator

Answer 6

oh hi

Answer 7

and hey

Answer 8

you will have to know how to factor a difference of cubes

Answer 9

but using u sustitution is easier

Answer 10

\[\lim_{u \rightarrow 2} \frac{ \sqrt{7+u}-3}{u^3-8}\]

Answer 11

if you replace cube root of x with u
you have to replace x with the cube of u

Answer 12

oh i see

Answer 13

and also cahnge the x ---> 8 to u ----> 2 ?

Answer 14

oh i see

Answer 15

\[\lim_{x \rightarrow 8} \sqrt[3]{x}=2\]
since u equals the cube root of x

Answer 16

hmm i see

Answer 17

you also need to rationalize the numerator

Answer 18

to get something to cancel

Answer 19

ohhh yeah thanks

Answer 20

yeah i guess that means you got it :)

Answer 21

something disappeared

Answer 22

well in the end i end up with 1 / u^2 + u + 4

Answer 23

so do i just repalce u with 2 and get the limit

Answer 24

\[\lim_{u \rightarrow 2}\frac{\sqrt{7+u}-3}{u^3-8} \\ \lim_{x \rightarrow 2}\frac{7+u-9}{(u^3-8)(\sqrt{7+u}+3)} \\ \lim_{u \rightarrow 2}\frac{1}{(u^2+2u+4)(\sqrt{7+u}+3)}\]

Answer 25

omg hehehe i forgot that :|

Answer 26

Thank You ~~~~

Answer 27

np